Question:

Step 1: The given condition states that $a, b, c$ are three distinct non-zero real numbers satisfying the equation $\dfrac{1}{x} + \dfrac{1}{x-1} + \dfrac{1}{x-2} = 1$. To find the polynomial whose roots are $a, b, c$, we clear the denominators: $$ \frac{1}{x} + \frac{1}{x-1} + \frac{1}{x-2} = 1 $$ Multiply both sides by $x(x-1)(x-2)$: $$ (x-1)(x-2) + x(x-2) + x(x-1) = x(x-1)(x-2) $$ Step 2: Expand and simplify the equation. The left side expands to: $$ (x^2-3x+2) + (x^2-2x) + (x^2-x) = 3x^2-6x+2 $$ The right side expands to: $$ x(x^2-3x+2) = x^3-3x^2+2x $$ Equating both sides, we obtain: $$ 3x^2-6x+2 = x^3-3x^2+2x $$ Rearranging the terms to form a standard cubic polynomial equation: $$ x^3-6x^2+8x-2=0 $$ The roots of this cubic equation are $a, b, c$. Step 3: Apply Vieta's formulas to determine the product $abc$. For a cubic equation $Ax^3+Bx^2+Cx+D=0$, the product of the roots is given by $-D/A$. In our equation $x^3-6x^2+8x-2=0$, we have $A=1$, $B=-6$, $C=8$, and $D=-2$. Therefore, the product of the roots $a, b, c$ is: $$ abc = -\frac{-2}{1} = 2 $$ Step 4: Calculate the value of $(1-a)(1-b)(1-c)$. The polynomial whose roots are $a, b, c$ is $P(x) = x^3-6x^2+8x-2$. This polynomial can also be expressed in factored form as $P(x) = (x-a)(x-b)(x-c)$. To find $(1-a)(1-b)(1-c)$, we evaluate $P(x)$ at $x=1$: $$ (1-a)(1-b)(1-c) = P(1) $$ Substitute $x=1$ into the polynomial: $$ P(1) = (1)^3 - 6(1)^2 + 8(1) - 2 $$ $$ P(1) = 1 - 6 + 8 - 2 $$ $$ P(1) = 1 $$ Thus, $(1-a)(1-b)(1-c) = 1$.